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            <h1 id="seo-header">『算法-ACM竞赛-最短路』Floyd —Warshall（最短路及其他用法详解）</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-最短路』Floyd-—Warshall（最短路及其他用法详解）"><a href="#『算法-ACM-竞赛-最短路』Floyd-—Warshall（最短路及其他用法详解）" class="headerlink" title="『算法-ACM 竞赛-最短路』Floyd —Warshall（最短路及其他用法详解）"></a>『算法-ACM 竞赛-最短路』Floyd —Warshall（最短路及其他用法详解）</h1><h3 id="一、多元最短路求法"><a href="#一、多元最短路求法" class="headerlink" title="一、多元最短路求法"></a>一、多元最短路求法</h3><p>多元都求出来了，单源的肯定也能求。<br>思想是动态规划的思想：从任意节点 A 到任意节点 B 的最短路径不外乎 2 种可能，1 是直接从 A 到 B，2 是从 A 经过若干个节点 X 到 B。所以，我们假设 Dis(AB)为节点 A 到节点 B 的最短路径的距离，对于每一个节点 X，我们易写出状态转移方程 Dis(AB) &#x3D;min（Dis(AX) + Dis(XB) ，Dis(AB)）这样一来，当我们遍历完所有节点 X，Dis(AB)中记录的便是 A 到 B 的最短路径的距离。</p>
<p>memset（Dis，0x3f,sizeof(Dis);<br>&#x2F;&#x2F;初始化，这里采用 0x3f 而非 0x7f，是当两个 0x7f7f7f7f 相加符号变号成为一个无穷小量。<br>void floyd(int N）<br>{<br>int i,j,k;<br>for(k&#x3D;0;k&lt;N;k++)<br>{<br>for(i&#x3D;0;i&lt;N;i++)<br>{<br>for(j&#x3D;0;j&lt;N;j++)<br>{<br>if(Dis[i][k]+Dis[k][j]&lt;Dis[i][j])<br>{<br>Dis[i][j]&#x3D;Dis[i][k]+Dis[k][j];</p>
<pre><code class="hljs">                &#125;
            &#125;
        &#125;
    &#125;
&#125;
</code></pre>
<p>这里一定要把 K 写到外边，需要先更新 K 前面的点在更新 K 后的点才有意义。<br><img src="https://img-blog.csdnimg.cn/20190521001639964.gif" srcset="/img/loading.gif" lazyload alt="在这里插入图片描述"></p>
<blockquote>
<p>结合代码 并参照上图所示 我们来模拟执行下 这样才能加深理解：<br>第一关键步骤：当 k 执行到 x，i&#x3D;v,j&#x3D;u 时，计算出 v 到 u 的最短路径要通过 x，此时 v、u 联通了。<br>第二关键步骤：当 k 执行到 u，i&#x3D;v，j&#x3D;y，此时计算出 v 到 y 的最短路径的最短路径为 v 到 u，再到 y(此时 v 到 u 的最短路径上一步我们已经计算过来，直接利用上步结果)。<br>第三关键步骤：当 k 执行到 y 时，i&#x3D;v，j&#x3D;w，此时计算出最短路径为 v 到 y(此时 v 到 y 的最短路径长在第二步我们已经计算出来了)，再从 y 到 w。<br>依次扫描每一点(k)，并以该点作为中介点，计算出通过 k 点的其他任意两点(i,j)的最短距离，这就是 floyd 算法的精髓！同时也解释了为什么 k 点这个中介点要放在最外层循环的原因.</p>
</blockquote>
<p>完整代码：</p>
<figure class="highlight prolog"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br></pre></td><td class="code"><pre><code class="hljs prolog">#include&lt;iostream&gt;<br>#include&lt;stack&gt;<br>using namespace std;<br>#define <span class="hljs-symbol">MAX</span> <span class="hljs-number">1000</span><br>int <span class="hljs-symbol">Graph</span>[<span class="hljs-symbol">MAX</span>][<span class="hljs-symbol">MAX</span>];<br>int <span class="hljs-symbol">Dis</span>[<span class="hljs-symbol">MAX</span>][<span class="hljs-symbol">MAX</span>];<br>#define infinite <span class="hljs-number">1000</span><br>int path[<span class="hljs-symbol">MAX</span>][<span class="hljs-symbol">MAX</span>];<br><br>void floyd(int <span class="hljs-symbol">N</span>)<br>&#123;<br>	int i,j,k;<br>	for(k=<span class="hljs-number">0</span>;k&lt;<span class="hljs-symbol">N</span>;k++)<br>	&#123;<br>		for(i=<span class="hljs-number">0</span>;i&lt;<span class="hljs-symbol">N</span>;i++)<br>		&#123;<br>			for(j=<span class="hljs-number">0</span>;j&lt;<span class="hljs-symbol">N</span>;j++)<br>			&#123;<br>				if(<span class="hljs-symbol">Dis</span>[i][k]+<span class="hljs-symbol">Dis</span>[k][j]&lt;<span class="hljs-symbol">Dis</span>[i][j])<br>				&#123;<br>					<span class="hljs-symbol">Dis</span>[i][j]=<span class="hljs-symbol">Dis</span>[i][k]+<span class="hljs-symbol">Dis</span>[k][j];<br>					path[i][j]=k;<br><br>				&#125;<br>			&#125;<br>		&#125;<br>	&#125;<br><br>&#125;<br><br>void print_path(int <span class="hljs-symbol">N</span>)<br>&#123;<br>	int i,j;<br>	for(i=<span class="hljs-number">0</span>;i&lt;<span class="hljs-symbol">N</span>;i++)<br>	&#123;<br>		for(j=<span class="hljs-number">0</span>;j&lt;<span class="hljs-symbol">N</span>;j++)<br>		&#123;<br>			if((i!=j) &amp;&amp;<span class="hljs-symbol">Dis</span>[i][j]!=infinite)<br>			&#123;<br>				cout&lt;&lt;i+<span class="hljs-number">1</span>&lt;&lt;<span class="hljs-string">&quot;----&quot;</span>&lt;&lt;j+<span class="hljs-number">1</span>&lt;&lt;<span class="hljs-string">&quot;   distance:&quot;</span>&lt;&lt;<span class="hljs-symbol">Dis</span>[i][j]&lt;&lt;endl;<br>				cout&lt;&lt;<span class="hljs-string">&quot;path:&quot;</span>&lt;&lt;endl;<br>				int k=j;<br>				stack &lt;int&gt; ph;<br>				do<br>				&#123;<br>					k=path[i][k];<br>					ph.push(k);<br>				&#125;while(k!=i);<br>				cout&lt;&lt;ph.top()+<span class="hljs-number">1</span>;<br>				ph.pop();<br>				while(!ph.empty())<br>				&#123;<br>					cout&lt;&lt;<span class="hljs-string">&quot;-&gt;&quot;</span>&lt;&lt;ph.top()+<span class="hljs-number">1</span>;<br>					ph.pop();<br>				&#125;<br>				cout&lt;&lt;<span class="hljs-string">&quot;-&gt;&quot;</span>&lt;&lt;j+<span class="hljs-number">1</span>&lt;&lt;endl;<br>			&#125;<br>		&#125;<br>	&#125;<br>&#125;<br><br>void main()<br>&#123;<br>	int <span class="hljs-symbol">N</span>,i,j;<br>	cin&gt;&gt;<span class="hljs-symbol">N</span>;<br>	for(i=<span class="hljs-number">0</span>;i&lt;<span class="hljs-symbol">N</span>;i++)<br>	&#123;<br>		for(j=<span class="hljs-number">0</span>;j&lt;<span class="hljs-symbol">N</span>;j++)<br>		&#123;<br>			int g;<br>			cin&gt;&gt;g;<br>			<span class="hljs-symbol">Graph</span>[i][j]=g;<br>			<span class="hljs-symbol">Dis</span>[i][j]=g;<br>		&#125;<br>	&#125;<br>//初始化路径<br>		for(i=<span class="hljs-number">0</span>;i&lt;<span class="hljs-symbol">N</span>;i++)<br>		&#123;<br>			for(j=<span class="hljs-number">0</span>;j&lt;<span class="hljs-symbol">N</span>;j++)<br>			&#123;<br>				path[i][j]=i;<br>			&#125;<br>		&#125;<br>	floyd(<span class="hljs-symbol">N</span>);<br>	print_path(<span class="hljs-symbol">N</span>);<br>    system(<span class="hljs-string">&quot;pause&quot;</span>);<br>&#125;<br></code></pre></td></tr></table></figure>

<h3 id="二、连通性"><a href="#二、连通性" class="headerlink" title="二、连通性"></a>二、连通性</h3><p>讲 Dis[i][j]不连联通时设置为 0，联通时设置为 1.<br>则可得状态转移方程<br>dis[i][j]&#x3D;dp[i][j]||(dp[i][k]&amp;&amp;dp[k][j]);<br>跟上面代码除了状态转移方程之外还有初始化不同，这个都初始化为 0；<br>其余都一样。要么 ij 直接连通，要么 ij 通过 K 联通。</p>
<figure class="highlight inform7"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br></pre></td><td class="code"><pre><code class="hljs inform7">void floyd(int N)<br>&#123;<br>	int i,j,k;<br>	for(k=0;k&lt;N;k++)<br>	&#123;<br>		for(i=0;i&lt;N;i++)<br>		&#123;<br>			for(j=0;j&lt;N;j++)<br>			&#123;<br>				if((dp<span class="hljs-comment">[i]</span><span class="hljs-comment">[k]</span>&amp;&amp;dp<span class="hljs-comment">[k]</span><span class="hljs-comment">[j]</span>)&amp;&amp;!Dis<span class="hljs-comment">[i]</span><span class="hljs-comment">[j]</span>)<br>				&#123;<br>					Dis<span class="hljs-comment">[i]</span><span class="hljs-comment">[j]</span>=Dis<span class="hljs-comment">[i]</span><span class="hljs-comment">[k]</span>+Dis<span class="hljs-comment">[k]</span><span class="hljs-comment">[j]</span>;<br>					path<span class="hljs-comment">[i]</span><span class="hljs-comment">[j]</span>=k;<br>				&#125;<br>			&#125;<br>		&#125;<br>	&#125;<br>&#125;<br></code></pre></td></tr></table></figure>

<h3 id="三、求无向图中可以删除一些边，使得任意两点的最短路不改变，求这些边能删除的最大的条数。（最小生成树问题）"><a href="#三、求无向图中可以删除一些边，使得任意两点的最短路不改变，求这些边能删除的最大的条数。（最小生成树问题）" class="headerlink" title="三、求无向图中可以删除一些边，使得任意两点的最短路不改变，求这些边能删除的最大的条数。（最小生成树问题）"></a>三、求无向图中可以删除一些边，使得任意两点的最短路不改变，求这些边能删除的最大的条数。（最小生成树问题）</h3><p>首先先在输入边的时候将重边去掉，保留最小的。<br>然后进行佛洛依德。<br>如果原来两点的最短距离大于经过第三个点的最短距离的话，那么我们就将这两点的最短距离<br>替换成经过第三条边的最短距离，当循环节结束后通过对比两点之间的距离变化，即可知哪些边将被删去。但是~~~当两点之间本来没有边的情况下，我们肯定是经过第三个点所到达的。那么就没有替换原来的边，这种情况的话，就直接 continue；</p>
<h3 id="四、无向图最小环"><a href="#四、无向图最小环" class="headerlink" title="四、无向图最小环"></a>四、无向图最小环</h3><p>若用 dis[i][j]表示 ij 之间的最小值，则由 i j 加线外一点 k 的环值为 dis[i][j]+length[i][k]+length[k][j];<br>枚举中间点 k，在用其更新最短路前，先找最小环，令 1&lt;&#x3D;i&lt;j&lt;k，即 k 点必定不在 i,j 的最短路上，则这个环中至少有三个点，可得状态转移方程 ans&#x3D;min(ans,dis[i][j]+length[i][k]+length[k][j]);</p>
<pre><code class="hljs">#include &lt;cstdio&gt;
#include &lt;cstring&gt;
#include &lt;map&gt;
#include &lt;queue&gt;
#include &lt;algorithm&gt;

using namespace std;

struct Node &#123;
    int s[9];//s数组表示包括本端所连的fence

    Node() &#123;
        memset(s,0,sizeof(s));
    &#125;

    bool operator &lt; (const Node&amp; a) const &#123;
    for(int i=0;i&lt;9;++i)
        if(s[i]&lt;a.s[i])
            return true;
        else if(s[i]&gt;a.s[i])
            return false;
    return false;
&#125;

bool operator ==(const Node&amp; a) const &#123;
    for(int i=0;i&lt;9;++i)
        if(s[i]!=a.s[i])
            return false;
    return true;
&#125;

&#125;fence[205];
int n,s,ls,ns,n1s,n2s,sta,des,cur;
int g[105][105],cnt=0,dis[105][105];
bool vis[105];
map&lt;Node,int&gt; mp;
int floyd() &#123;
    int ans=0x1f1f1f1f;
    for(int i=1;i&lt;=n;++i)
        for(int j=i;j&lt;=n;++j)
            dis[i][j]=dis[j][i]=g[i][j];

    for(int k=1;k&lt;=cnt;++k) &#123;
        for(int i=1;i&lt;k;++i)//寻找最小环
        for(int j=i+1;j&lt;k;++j)
            if(dis[i][j]+g[i][k]+g[k][j]&lt;ans)//由于此处会存在三个INF相加，所以INF设为0x1f1f1f1f
                ans=dis[i][j]+g[i][k]+g[k][j];
    for(int i=1;i&lt;=n;++i)//更新最短路
        for(int j=1;j&lt;=n;++j)
            if(dis[i][j]&gt;dis[i][k]+dis[k][j])
                dis[i][j]=dis[i][k]+dis[k][j];
&#125;
return ans;
&#125;
int main() &#123;
    //freopen(&quot;fence6.in&quot;,&quot;r&quot;,stdin);
   // freopen(&quot;fence6.out&quot;,&quot;w&quot;,stdout);

    memset(g,0x1f,sizeof(g));
    scanf(&quot;%d&quot;,&amp;n);
    for(int i=1;i&lt;=n;++i) &#123;//读入边数据，并给每个点标一个数
        scanf(&quot;%d%d%d%d&quot;,&amp;s,&amp;ls,&amp;n1s,&amp;n2s);
        fence[i&lt;&lt;1].s[8]=fence[(i&lt;&lt;1)|1].s[8]=s;

    while(n1s--&gt;0)
        scanf(&quot;%d&quot;,&amp;fence[i&lt;&lt;1].s[n1s]);
    sort(fence[i&lt;&lt;1].s,fence[i&lt;&lt;1].s+9);
    if(mp[fence[i&lt;&lt;1]]==0)
        mp[fence[i&lt;&lt;1]]=++cnt;

    while(n2s--&gt;0)
        scanf(&quot;%d&quot;,&amp;fence[(i&lt;&lt;1)|1].s[n2s]);
    sort(fence[(i&lt;&lt;1)|1].s,fence[(i&lt;&lt;1)|1].s+9);
    if(mp[fence[(i&lt;&lt;1)|1]]==0)
        mp[fence[(i&lt;&lt;1)|1]]=++cnt;

    sta=mp[fence[i&lt;&lt;1]];
    des=mp[fence[(i&lt;&lt;1)|1]];
    g[sta][des]=g[des][sta]=ls;//边信息转成点信息
&#125;
printf(&quot;%d\n&quot;,floyd());
return 0;
&#125;
</code></pre>
<h3 id="五、传递闭包问题"><a href="#五、传递闭包问题" class="headerlink" title="五、传递闭包问题"></a>五、传递闭包问题</h3><p>邻接矩阵是显示两点的直接关系，如 a 直接能到 b，就为 1。而传递闭包显示的是传递关系，如 a 不能直接到 c，却可以通过 a 到 b 到 d 再到 c，因此 a 到 c 为 1。<br><img src="https://img-blog.csdn.net/20170409190428811?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvd3p3MTM3NjEyNDA2MQ==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" srcset="/img/loading.gif" lazyload alt="在这里插入图片描述"><br>另外矩阵 A 进行自乘即 A^{2}得到的矩阵中，为 1 的值表示走最多两步可以到达。A^{3}矩阵中为 1 的值表示，最多走三步可以到达。<br>简单来说，就是有向图确定先后顺序。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br></pre></td><td class="code"><pre><code class="hljs cpp"><span class="hljs-comment">/*</span><br><span class="hljs-comment">题目：n头牛进行m场比赛，问能确定排名的有多少头牛。</span><br><span class="hljs-comment">  解答：构造一个n个点的有向图，如果牛a胜b，那么a-&gt;b，如果a-&gt;b，b-&gt;c，则有a-&gt;c，这个用floyd。</span><br><span class="hljs-comment">  最后得到该图的传递闭包link的二维数组。最后统计每一个点入度和出度和为n-1的点的个数即可。</span><br><span class="hljs-comment">*/</span><br><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;stdio.h&gt;</span></span><br><span class="hljs-meta">#<span class="hljs-keyword">include</span><span class="hljs-string">&lt;string.h&gt;</span></span><br><span class="hljs-type">const</span> <span class="hljs-type">int</span> MAX=<span class="hljs-number">105</span>;<br><span class="hljs-comment">/*</span><br><span class="hljs-comment">有向图的传递闭包！</span><br><span class="hljs-comment">注意传递之前一定要初始化！</span><br><span class="hljs-comment">如果i!=j&amp;&amp;(i,j)不属于E(边的集合) t[i][j]=0;</span><br><span class="hljs-comment">如果i=j||(i,j)属于E(边的集合)     t[i][j]=1;</span><br><span class="hljs-comment">*/</span><br><br><span class="hljs-comment">//传递闭包</span><br><span class="hljs-function"><span class="hljs-type">void</span> <span class="hljs-title">Transitive_Closure</span><span class="hljs-params">(<span class="hljs-type">int</span> n,<span class="hljs-type">bool</span> t[][MAX])</span></span><br><span class="hljs-function"></span>&#123;<br>	<span class="hljs-type">int</span> i,j,k;<br>	<span class="hljs-keyword">for</span>(k=<span class="hljs-number">1</span>;k&lt;=n;k++)<br>		<span class="hljs-keyword">for</span>(i=<span class="hljs-number">1</span>;i&lt;=n;i++)<br>			<span class="hljs-keyword">for</span>(j=<span class="hljs-number">1</span>;j&lt;=n;j++)<br>				t[i][j]=t[i][j]|(t[i][k]&amp;t[k][j]);<br>&#125;<br><span class="hljs-function"><span class="hljs-type">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span><br><span class="hljs-function"></span>&#123;<br>	<span class="hljs-type">int</span> n,i,j,m,st,ed,sum,num;<br>	<span class="hljs-type">bool</span> t[MAX][MAX];<br>	<span class="hljs-keyword">while</span>(<span class="hljs-built_in">scanf</span>(<span class="hljs-string">&quot;%d%d&quot;</span>,&amp;n,&amp;m))<br>	&#123;<br>		<span class="hljs-keyword">if</span>(n==<span class="hljs-number">0</span>&amp;&amp;m==<span class="hljs-number">0</span>)<br>			<span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;<br>		<span class="hljs-built_in">memset</span>(t,<span class="hljs-literal">false</span>,<span class="hljs-built_in">sizeof</span>(t));<br>		<span class="hljs-keyword">for</span>(i=<span class="hljs-number">1</span>;i&lt;=n;i++)<br>			t[i][i]=<span class="hljs-literal">true</span>;<br>		<span class="hljs-keyword">for</span>(i=<span class="hljs-number">1</span>;i&lt;=m;i++)<br>		&#123;<br>			<span class="hljs-built_in">scanf</span>(<span class="hljs-string">&quot;%d%d&quot;</span>,&amp;st,&amp;ed);<br>			t[st][ed]=<span class="hljs-literal">true</span>;<br>		&#125;<span class="hljs-comment">//上面的代码都是初始化</span><br>		<span class="hljs-built_in">Transitive_Closure</span>(n,t);<br>		sum=<span class="hljs-number">0</span>;<br>		<span class="hljs-keyword">for</span>(i=<span class="hljs-number">1</span>;i&lt;=n;i++)<br>		&#123;<br>			num=<span class="hljs-number">0</span>;<br>			<span class="hljs-keyword">for</span>(j=<span class="hljs-number">1</span>;j&lt;=n;j++)<br>				<span class="hljs-keyword">if</span>(i==j)<br>					<span class="hljs-keyword">continue</span>;<br>				<span class="hljs-keyword">else</span><br>					num+=(t[i][j]||t[j][i]);<span class="hljs-comment">//统计出度和入度的个数!</span><br>				sum+=(num==n<span class="hljs-number">-1</span>);<br>		&#125;<br>		<span class="hljs-built_in">printf</span>(<span class="hljs-string">&quot;%d\n&quot;</span>,sum);<br>	&#125;<br>	<span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;<br>&#125;<br><span class="hljs-comment">/*</span><br><span class="hljs-comment">5 5</span><br><span class="hljs-comment">4 3</span><br><span class="hljs-comment">4 2</span><br><span class="hljs-comment">3 2</span><br><span class="hljs-comment">1 2</span><br><span class="hljs-comment">2 5</span><br><span class="hljs-comment">  2</span><br><span class="hljs-comment">*/</span><br></code></pre></td></tr></table></figure>

                
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